Simple Harmonic Motion

Question

Displacement where the kinetic energy and the potential energy are the
same in Simple Harmonic Motion.

 

I know We need to equate the 2 equations

 

1/2(mv²) = 1/2 kx²

 

I know the answer is supposed to be x  =A/√2

 

No idea how to do the maths in this question I know V max= Aw where w =√k/m. Tried messing around and can't get  the answer. I ended up with A² = x²

Answer 1

Hi Ian!

 

I'm sorry I don't have a lot of time right now, but I can suggest a solution for this one.

 

In this sort of simple harmonic motion, total mechanical energy is conserved.  The mechanical energy is all (elastic) potential, when the object is at its maximum amplitude (A), so we can write the total mechanical energy in terms of the amplitude as:

 

ME_tot = (1/2)kA²

 

For the position of interest, this total mechanical energy is evenly split between kinetic and (elastic) potential.  Thus, the potential energy at that position (x_1/2) is (1/2)ME_tot = (1/2)(1/2)kA²

 

Thus:

 

(1/2)kx_1/2² = (1/2)(1/2)kA²

 

If you solve that expression for x_1/2, I think you will get the answer you seek.  Just let me know if you have any questions, or want to talk about this further.